Question

The integral $\underset{-1/2}{\overset{1/2}{\int }}\{\left[x\right]+\log \left(\frac{1+x}{1-x}\right)\}dx$ is equal to $($ where $[\cdot ]$ represents greatest integer function$)$

• A. $-\frac{1}{2}$
• B. $0$
• C. $1$
• D. $2\log$ $\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

Let $I=\underset{-1/2}{\overset{1/2}{\int }}\{\left[x\right]+\log \left(\frac{1+x}{1-x}\right)\}dx$
$=\underset{-1/2}{\overset{1/2}{\int }}\left[x\right]dx+\underset{-1/2}{\overset{1/2}{\int }}\log \left(\frac{1+x}{1-x}\right)dx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$\Rightarrow I=\underset{-1/2}{\overset{1/2}{\int }}\left[x\right]dx+0,(\because \log \left(\frac{1+x}{1-x}\right)\text{is an odd function})$
$=\underset{-1/2}{\overset{0}{\int }}\left[x\right]dx+\underset{0}{\overset{1/2}{\int }}\left[x\right]dx$
$=\underset{-1/2}{\overset{0}{\int }}-1dx+\underset{0}{\overset{1/2}{\int }}0dx$
$={[-x]}_{-1/2}^0=-\frac{1}{2}$